1.  Verify that 2^35 -1 is divisible by 127

We are asked to verify that \( 2^{35} - 1 \) is divisible by 127. In other words, we need to show that \( 2^{35} - 1 \equiv 0 \mod 127 \), which implies that \( 2^{35} \equiv 1 \mod 127 \).

### Step 1: Use Fermat's Little Theorem
By Fermat's Little Theorem, if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

\[
a^{p-1} \equiv 1 \mod p
\]

In our case, \( p = 127 \), which is a prime number, and \( a = 2 \). Therefore, Fermat's Little Theorem tells us:

\[
2^{126} \equiv 1 \mod 127
\]

This means that the order of 2 modulo 127 must divide 126. The order is the smallest integer \( d \) such that \( 2^d \equiv 1 \mod 127 \).

### Step 2: Find the Order of 2 Modulo 127
Let's calculate the order of 2 modulo 127 by trying smaller exponents. We seek the smallest \( d \) such that \( 2^d \equiv 1 \mod 127 \).

1. **Try \( d = 7 \):**

   Calculate \( 2^7 \mod 127 \):

   \[
   2^7 = 128 \equiv 1 \mod 127
   \]

So, \( 2^7 \equiv 1 \mod 127 \). This tells us that the order of 2 modulo 127 is 7.

### Step 3: Check if \( 35 \) is a Multiple of the Order
Since the order of 2 modulo 127 is 7, any power of 2 that is a multiple of 7 will be congruent to 1 modulo 127. Specifically, we have:

\[
35 \div 7 = 5
\]

This shows that \( 2^{35} = (2^7)^5 \). Since \( 2^7 \equiv 1 \mod 127 \), it follows that:

\[
2^{35} \equiv 1^5 \equiv 1 \mod 127
\]

### Step 4: Conclude the Proof
Since \( 2^{35} \equiv 1 \mod 127 \), it follows that:

\[
2^{35} - 1 \equiv 0 \mod 127
\]

Thus, \( 2^{35} - 1 \) is divisible by 127, as required.

---

Do you have any questions or need further explanation? Here are five related questions to explore:

1. Can you use Fermat's Little Theorem to check if \( 2^{126} - 1 \) is divisible by 127?
2. How would you find the order of a number modulo a different prime number?
3. Can you use the Chinese Remainder Theorem to verify similar divisibility problems?
4. What is the significance of primitive roots in modular arithmetic?
5. How does Lagrange's theorem help in solving modular arithmetic problems?

**Tip:** When dealing with large powers in modular arithmetic, consider using theorems like Fermat's Little Theorem to simplify calculations.

Learn how to verify divisibility of 2 raised to the power of 35 minus 1 by 127 using Fermat's Little Theorem and modular arithmetic. Detailed step-by-step explanation with calculations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation